Question

Let $f(x)=x^2\log x,x>0$. Then the minimum value of $f$ is

• A. $\frac{1}{\sqrt{e}}$
• B. $2e$
• C. $-2e$
• D. $\sqrt{e}$
• E. $\frac{-1}{2e}$

Answer 1

Answer: (E)

$f(x)=x^2\log x,x>0$
$f^{\prime }(x)=x^2\times \frac{1}{x}+\log x\times 2x$
$=x(1+2\log x)$
$=x\left(1+\log x^2\right)$
$f^{\prime }(x)=0$
As $x\ne 0,1+\log x^2=0$
$\log x^2=-1$
$x^2=e^{-1}$
$x=\frac{1}{\sqrt{e}}$
$f^{\prime \prime }(x)=1+\frac{2}{x}$
$f^{\prime \prime }\left(\frac{1}{\sqrt{6}}\right)=1+2\sqrt{e}>0$
$\text{ Hence, }f(x)\text{ is minimum at }x=\frac{1}{\sqrt{6}}$
$f\left(\frac{1}{\sqrt{e}}\right)=\frac{1}{e}\log e^{-1/2}=\frac{-1}{2e}$